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Find a number field whose unit group is isomorphic to $\mathbb{Z}/4\mathbb{Z} \times \mathbb{Z}.$

I'm trying to use Dirichlet's Unit Theorem to solve this problem. It states that if $K$ is a number field of signature $(r,s)$ and $\mu_K$ is the set of roots of unity in $K$, then the unit group $\mathcal{O}_K^{\times}$ of the ring of integers is isomorphic to $\mu_K \times \mathbb{Z}^{r+s−1}$ as an abelian group. So I suppose I want $r+s-1=1,$ or $r+s=2$. This forces $(r,s)=(0,2)$ because if there's at least one real embedding then $\mu_K$ is just $\{\pm 1\}$ so not $\mathbb{Z}/4\mathbb{Z}$.

Therefore I need a number field of degree $4$ with four complex embeddings and whose set of roots of unity is isomorphic to $\mathbb{Z}/4\mathbb{Z}$. The cyclotomic field $\mathbb{Q}(\zeta_5)$ doesn't work because it has more than $4$ elements in its set of roots of unity (and $\mathbb{Q}(\zeta_4)$ doesn't work because it doesn't have degree $4$). I suppose it will have to be $\mathbb{Q}(\alpha)$ where the minimal polynomial of $\alpha$ has degree $4$ but I haven't been able to find an example. Hints would be appreciated.

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    $\begingroup$ If the set of roots of unity in your $K$ has to be cyclic of order $4$ then it has to be $\{\pm1,\pm i\}$ under any complex embedding. Thus, I would search your $K$ among those of the form $\Bbb Q(i,\sqrt{d})$ with $d<0$ to start with. $\endgroup$ – Andrea Mori Mar 15 at 20:35
  • $\begingroup$ So perhaps I could take $\mathbb{Q}(i, \sqrt{2})$? That has degree $4$ and four complex embeddings. Its set of roots of unity includes $\pm 1,\pm i$, so it would suffice to show there are no other roots of unity in this set. Any other root of unity would need to be a primitive $m$th root of unity with $m=3$ or $m \geq 5$. I'm not sure how to show this is impossible though. $\endgroup$ – AlephNull Mar 15 at 21:12
  • $\begingroup$ Does $\Bbb Q(\zeta_8)$ get you where you want to be? $\endgroup$ – Robert Shore Mar 15 at 22:21
  • $\begingroup$ @RobertShore I considered that but I'm not sure what made me refute it. It certainly has degree $4$ and four complex embeddings. But doesn't that have more than $4$ roots of unity? $\endgroup$ – AlephNull Mar 15 at 22:50
  • $\begingroup$ @AlephNull I think your example of $\mathbb{Q}(i, \sqrt{2})$ works. You can check that the roots of unity in $\mathbb{Q}(\zeta_n)$ are $\mu_n$ if $n$ is even and $\mu_{2n}$ if $n$ is odd. (Intuitively, this is because multiplying by $-1$ will double the order of an odd root of unity) $\endgroup$ – vxnture Mar 15 at 23:00
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You are almost there! Since $K$ has a torsion element of order $4$, it contains $\zeta_4$ and thus contains $\mathbf{Q}(\zeta_4)$. Dirichlet's unit theorem then says that $K$ must be degree $4$ and thus a quadratic extension of $\mathbf{Q}(\zeta_4)$.

Now suppose that $K$ is any quadratic extension of $\mathbf{Q}(\zeta_4) = \mathbf{Q}(\sqrt{-1})$. The signature of $K$ is $(0,2)$ and so $K$ has unit rank one. Also, $K$ has an element $\zeta_4$ of order $4$. The only thing that remains is to find a $K$ that doesn't have any extra torsion. But the torsion subgroup of a number field is always cyclic and generated by a $m$th root of unity, or a $4n$th root of unity in this case since we already have a $4$th root of unity. So you just have to ensure that

$$\mathbf{Q}(\zeta_{4n}) \not\subset K$$

for any $n > 1$. The degree of $\mathbf{Q}(\zeta_{4n})$ is (Euler's $\varphi$ function) $\varphi(4n)$. This is $> 4$ for $n \ge 4$. So the answer is:

$K$ can be any quadratic extension of $\mathbf{Q}(\zeta_4)$ which doesn't equal $\mathbf{Q}(\zeta_8)$ or $\mathbf{Q}(\zeta_{12})$.

Since $\mathbf{Q}(\zeta_8) = \mathbf{Q}(\sqrt{-1},\sqrt{2}) = \mathbf{Q}(\sqrt{-1},\sqrt{-2})$ and $\mathbf{Q}(\zeta_{12}) = \mathbf{Q}(\sqrt{-1},\sqrt{-3}) = \mathbf{Q}(\sqrt{-1},\sqrt{2})$, you can find many such $K$, for example $K = \mathbf{Q}(\sqrt{-1},\sqrt{d})$ for any squarefree $\pm d > 3$. These are not the only examples --- the others are precisely all the quadratic extensions $K/\mathbf{Q}(\sqrt{-1})$ which are non-Galois over $\mathbf{Q}$ such as $\mathbf{Q}(i,\sqrt{3 + 4 i})$.

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  • $\begingroup$ Perfect answer, and this only uses results proved in my course. $\endgroup$ – AlephNull Mar 18 at 19:21

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